Name___________________________
Section__________________
7.014 Problem Set 1
Please print out this problem set and record your answers on the printed copy. Answers to this
problem set are to be turned in to the box outside 68-120 by 5:00pm on Friday February 16, 2007.
Problem sets will not be accepted late. Solutions will be posted online.
1. You are spending the summer working at the Venter Institute analyzing some of the new species collected
during the voyage of the Sorcerer II. You are working on three single-celled organisms, currently identified
as Species M, Species I and Species T. To begin characterizing these organisms, you want to determine how
they obtain energy.
(a) You first test for whether the organisms are heterotrophic or autotrophic. To do this, you grow all three
species under different conditions. You first grow them in liquid media with light in a closed system that
contains oxygen, nitrogen, phosphorous, all the salts in seawater, glucose and trace minerals. You also grow
the three species in liquid media with light in a closed system that contains the same mixture described
above, but this one has no glucose and is bubbled with CO2. You measure the growth of the three species
over time by counting the number of cells at specific time points:
With Glucose
Concentration of Cells
1.4
1.2
1
M
0.8
I
0.6
T
0.4
0.2
0
0
2
4
6
8
Time
With CO2
Concentration of Cells
1.2
1
0.8
M
0.6
I
0.4
T
0.2
0
0
2
4
6
8
Time
Which of the species are heterotrophs and which are autotrophs? Explain how the data above helped you
reach your conclusion.
Species I is a heterotroph and Species M and T are autotrophs. Species I is a heterotroph because it can not
grow in the absence of a fixed carbon source (without glucose), but it can grow when glucose, a source of
fixed carbon, is provided. Species M and T, however, can use CO2 as source of carbon to create a fixed
carbon source. Thus, Species M and T are autotrophs. Note that Species M and T can also grow in the
absence of CO2 (as shown from the data in the first graph). These autotrophs can grow “heterotrophically”
meaning that they can import the glucose from the media to use as a source of carbon rather than create
Question 1 continued
their own glucose by photosynthesizing CO2. Not all autotrophs can use glucose from the media as source of
organic carbon, but those that do are called mixotrophs.
(b) You want to determine if the autotrophs are photoautotrophs or chemoautotrophs. To do this, you grow
all species in liquid media, in a closed system, containing oxygen, nitrogen (ammonia), carbon dioxide,
phosphorous, all the salts in seawater, and trace minerals (note that there is no glucose provided). You
perform the experiment twice, once with light and once without light.
Light
Concentration of Cells
1.2
1
0.8
M
0.6
I
0.4
T
0.2
0
0
2
4
6
8
Time
Dark
Concentration of Cells
1.4
1.2
1
M
0.8
I
0.6
T
0.4
0.2
0
0
2
4
6
8
Time
Are any of the species photoautotrophs or chemoautotrophs? If so, explain how the data above helped you
reach your conclusion.
In the previous experiments, we discovered that Species I is a heterotroph, so it is neither a photoautotroph
nor a chemoautotroph and does not grow in the absence of glucose. Species T can not grow in the absence of
light. Thus, it is a photoautotroph. Species M can grow in either the light or the dark, so it must be a
chemoautotroph. It does not rely on the light for energy to initiate photosynthesis. Instead, it can reduce the
ammonia provided.
(c) Assume that you have identified a photoautotroph. You want to know if it is oxygenic or anoxygenic.
Propose an experiment that would allow you to differentiate between these two possibilities.
As discussed in lecture, anoxygenic photoautotrophs are strict anaerobes. To differentiate between oxygenic
and anoxygenic, you need to test if your photoautotroph can grow in the presence of oxygen. You could grow
the species in media with nitrogen, phosphorous, all the salts in seawater, glucose and trace minerals and
bubble either O2 or H2S through the media. Species that grow in the media with O2 present are oxygenic and
species that grow in the media with H2S present are anoxygenic.
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Question 1 continued
Alternatively, you know that your photoautotroph can not be an anoxygenic photoautotroph because all of
the previous experiments were done in the presence of oxygen, which would be lethal to an anoxygenic
photoautotroph.
2. Energy metabolism and redox reactions
(a) “The decay food chain does not always function efficiently. Under certain circumstances it exhausts all
the available oxygen. Decay is then incomplete; its products include methane, alcohols, amines, hydrogen
sulfide and partially decomposed organic matter.”
(i) What are the two predominant products of decomposition when oxygen is available?
CO2 and H2O
(ii) Does the above situation (decay in the absence of oxygen) describe decomposition in a reducing
environment or an oxidizing environment? Why?
Reducing environment. The products when there is no oxygen are fully reduced molecules indicating that
decay in the absence of oxygen results in a reducing environment.
(b) Over most of the last three billion years, photosynthesis has outpaced respiration on a global scale, as
evidenced in part by the presence of forests fossil fuels and oxygen in our atmosphere.
(i) Name the two chemical products of photosynthesis, and specify which one is a reducing agent and which
one is an oxidizing agent.
O2 and glucose (fixed carbon) are produced. O2 is an oxidzing agent and glucose is a reducing agent.
(ii) Predict the long term effect on 1) plant life and 2) other organisms on earth if respiration consistently
outpaced photosynthesis on a global scale.
If respiration outpaces photosynthesis, eventually all plant life would be consumed by other organisms in the
pursuit of energy. In the absence of a renewable energy source, eventually all organisms would run out of
useful energy.
(c) A schematic of the chemical, electron and energy flows in aerobic respiration is shown below. Note that
this diagram does not reflect the exact stoichiometry of the reactions, just the flow of reactants, products and
electrons (sold lines). Energy flows are shown as dotted lines.
CO2
glucose
e-
NADH
+H+
NAD+
O2
e-
H2O
H+ gradient
ADP + Pi
ATP
3
Question 2 continued
Draw a similar diagram for the metabolism of glucose by fermentation such that the products are ethanol and
CO2.
ADP + Pi
ATP
acetaldehyde
NADH
+H+
NAD+
e-
glucose
acetaldehyde
e-
CO2 + ethanol
(d) A schematic of the chemical, electron, and energy flows for one of many possible chemosynthesis
reactions is shown below:
NO2-
e-
NO3-
O2
H2O
H+ gradient
NO3NO2-
e-
NADPH
+H+
NADP+
ATP ADP + Pi
CO2
e-
glucose
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Question 2 continued
(i) Does the reaction in the dashed box create energy or use energy? Explain.
The reaction in the dashed box creates energy. This can be determined by adding together the energy from
each of the half reactions.
NO2
NO3- + eE0 = -0.42 V
O2 + e-
H2O
E0 = 0.81 V
ΔE0 = + 0.39 V
(ii) Does the reaction in the solid box create energy or use energy? Explain.
The reaction in the solid box uses energy. This can be determined by adding together the energy from each
of the half reactions.
NO2-
NO3- + e-
E0 = -0.42 V
NADP+ + e-
NADPH
E0 = -0.32 V
ΔE0 = -0.74 V
3. Each of the macromolecules that we have discussed in class is made up of a type of building block and
various types of bonds.
(a) Fill in the chart below describing these properties for each of the macromolecules:
Building Block
Bond(s) Involved
Carbohydrates
monosaccharide
covalent, hydrogen
DNA
deoxyribonucleotides
covalent, hydrogen
RNA
ribonucleotides
covalent, hydrogen
Protein
amino acids
covalent, hydrogen, ionic,
hydrophobic
(b) When these building blocks are put together, a specific reaction occurs. What is the general name for
this type of reaction? What product(s) are produced in each reaction in addition to the macromolecule?
The reaction is called a dehydration reaction because water is produced from creating the covalent bond
between subunits (one molecule of water is “removed” from the starting components).
(c) DNA and RNA are both synthesized directionally, 5’ → 3’. To what do 5’ and 3’ refer?
DNA and RNA are formed by the covalent connection of nucleotides via the phosphate group of the incoming
nucleotide reacting with the hydroxyl group of the last nucleotide in the growing chain of nucleotides. The
phosphate group is attached to the 5’ C of the ribose of the nucleotide and the hydroxyl group is attached to
the 3’ C of the ribose of the nucleotide. Because the first nucleotide will always have a free, or non-reacted,
phosphate group, the first nucleotide represents the 5’ end of the nucleotide chain. The last nucleotide has a
free hydroxyl group, thus it represents the 3’ end of the chain.
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Question 3 continued
(d) Proteins are synthesized directionality: from the N-terminus to the C-terminus. To what do N-terminus
and C-terminus refer?
Amino acids are connected via a peptide bond that forms between the amine group on one amino acid and
the carboxyl group on the another amino acid. The amine group of the incoming amino acid reacts with the
carboxyl group of the last amino acid on the growing chain of amino acids. The first amino acid in the chain
will have a free amine group, thus it is called the N-terminus (N is for the nitrogen atom in the amine group).
The last amino acid will have a free carboxyl group, thus it is called the C-terminus.
(e) The building blocks of carbohydrates can be referred to as α or as β. To what do α and β refer?
α and β represent the relative position of the hydrogen and hydroxyl groups attached to a carbon in a
monosaccharide.
4. For each of the descriptions below, list what type of bond is being described, where you might expect to
find that type of bond (in a biologically relevant molecule) and rank the strength of that bond in relation to
the other types of bonds using 1, 2, 3 or 4 with 1 being the strongest bond.
(a) This type of bond is the result of two particles of opposite charges being attracted to one another.
Bond: Ionic
Molecule: protein
Strength: 2
(b) This type of bond is the result of the random fluctuations of charge within a molecule. These random
fluctuations are a result of the movement of electrons and cause some regions of the molecule to be slightly
negative and some regions to be slightly positive. This distribution of charge is known as a dipole. Two
molecules can be attracted to each other due to opposing dipoles.
Bond: van der Waals
Molecule: all molecules exhibit van der waals
Strength: 4
(c) This type of bond is the result of two atoms equally sharing electrons between them.
Bond: covalent
Molecule: DNA, RNA, protein, carbohydrate etc.
Strength: 1
(d) This type of bond is the result of unequal sharing of electrons within a molecule such that one region of
the molecule is more electronegative than the other. The more electronegative region can then interact with
the less electronegative region of another molecule.
Bond: hydrogen
Molecule: DNA, protein
Strength: 3
5. To answer the question below, you need to use the StarBiochem program, a java viewer for
macromolecules. To get access to and directions for this viewer, go to http://web.mit.edu/viz/7.01x. Click
on “The StarBiochem Viewer” to first download the viewer. Once you have done that, click on “7.014
Problems”. You can first do the StarBiochem Tutorial Problem to learn how to use the program. Then
choose to do Problem 1, which will be the same as the question written below. Make sure to write your
answers in the spaces provided below.
If you do not have a computer easily available, room 37-212 has an Athena PC cluster
that will run the program you need.
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For the 37-212 cluster schedule, see http://web.mit.edu/acis/labs/37-312.html
If you would like to do this problem in the presence of technical staff who are familiar with
this computer program, there will be staff present on Thursday 2/15/07 from 4-9pm in
26-152. You are welcome to stop by anytime during that interval to use one of the
computers in that room to do the problem, or to ask the staff questions about the program.
Question 5 continued
We are examining the structure of a specific segment of DNA (1N14).
(a) Each piece of DNA shown here consists of 13 nucleotides.
(i) List in order from 5’ to 3’ the 13 nucleotides in this segment of DNA. Make sure to mark which end is the
5’ end and which end is the 3’ end.
5’-GCTAAGGAAAGCC-3’
or
5’-GGCTTTCCTTAGC-3’
(ii) List the 13 matching nucleotides from the opposite strand in order from the 3’ to 5’ end. Make sure to
mark which end is the 5’ end and which end is the 3’ end.
3’-CGATTCCTTTCGG-5’
or
3’-CCGAAAGGAATCG-5’
(b) Bonded segments of DNA run antiparallel to each other.
(i) Examine the 3’ end of one of the strands. Does it end with a sugar or a phosphate? Why is it called 3’?
It ends with a sugar. It is called the 3’ end because the last nucleotide in the chain has a non-reacted
hydroxyl group, which is attached to the 3’ C of the ribose.
(ii) Examine the 5’ end of one of the strands. Does it end with a sugar or a phosphate? Why is it called 5’?
It ends with a phosphate. It is called the 5’ end because the first nucleotide in the chain has a non-reacted
phosphate group, which is attached to the 5’ C of the ribose.
(c) The nitrogenous base regions of the molecule interact. Zoom in on stacked nitrogenous bases.
(i) Are the rings stacked parallel or perpendicular to each other?
Parallel
(ii) How many rings (count both connected and separated rings individually) total can you see in each stack?
Is this number uniform throughout the entire molecule?
There are three rings per stack (the purine has two rings and the pyrimidine has one). Yes, this number is
uniform. To keep the distance between the two strands of the double helix consistent, each stack always
consists of one purine and one pyrimidine.
(iii) What type of bond is primarily responsible for maintaining the nitrogenous base contacts?
The bond responsible for maintaining the contact between two nitrogenous bases on opposite strands is the
hydrogen bond.
(d) The shape of DNA is called a double helix. Examine this molecule of DNA closely.
(i) What features of this molecule help it retain its uniform double helix shape?
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The interaction between the nitrogenous base pairs helps form the double helix. The negative charge of the
sugar-phosphate backbone reacts with the aqueous environment ensuring that the nitrogenous bases remain
on the inside of the helix.
(ii) List three aspects of the molecule that would be different if we were looking at a molecule of RNA.
1. there would be a hydroxyl group on the 2’ C
2. RNA is single-stranded
3. there would be uracil instead of thymine.
Standard Eº values (at 25° C and pH 7)
+
-
Half Reaction
1/2 O2 + 2 H + 2 e
Fe3+ + eNO3- + 6 H+ + 6 eNO3- + 2 H+ + 2 eNO2- + 8 H+ + 6 eCH3OH + 2 H+ + 2 e2 H+ + 2 eacetaldehyde+ 2 H+ + 2 eCO2 + 8 H+ + 8 eNAD+ + H+ + 2 eNADP+ + H+ + 2 eCO2 + 4 H+ + 4 e -
→
→
→
→
→
→
→
→
→
→
→
→
E0 (V)
H2O
Fe2+
1/2 N2 + 3 H2O
NO2- + H2O
NH4 + 2 H2O
CH4 + H2O
H2 (pH 0)
ethanol
CH4 + 2 H2O
NADH
NADPH
1/6 glucose + H2O
0.816
0.771
0.75
0.421
0.34
0.17
0.00
-0.197
-0.24
-0.320
-0.324
-0.43
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